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I am just wondering if there is a general method in solving the Diophantine equation

$x^2-ay+c=0$,

where $a$ and $c$ are positive integers. This question arose because I need to find a solution to the equation $125+x^2=22y$. I am so thinking that trial and error is not a suitable method to do.

Thanks for your help.

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  • $\begingroup$ Few observations which probably will help you: x is odd and y is positive. $\endgroup$
    – Юрій Ярош
    May 14, 2018 at 14:48
  • $\begingroup$ You may be interested in this page, which implements an algorithm and presents step by step solutions to second order diophantine equations $\endgroup$
    – Mathmo123
    May 14, 2018 at 17:46

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I will solve just one specific case:

For this case, we need to solve $x^2 \equiv -125 \pmod{22}$

We have $22|(x^2+125)\Rightarrow22|(x^2-7+132)\Rightarrow22|(x^2-7)$ because $132$ is divisible by $22$.

So now we need to solve $x^2 \equiv 7 \pmod{22}$

Trial and error is actually suitable, but only if you notice that because $x^2-7$ is divisible by $22$, $x$ must be an odd number. This will reduce the number of cases from $22$ to $11$. With $11$ cases, we can easily set up a table like this:

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Remainder of }x\div22 &1&3&5&7&9&11&13&15&17&19&21\ \\ \hline \text{Remainder of }x^2\div22 &1&9&3&5&15&11&15&5&3&9&1\\\hline \text{Remainder of }(x^2-7)\div22 &16&2&18&20&8&4&8&20&18&2&16\\\hline \end{array}

Based on the table above, there are no positive integers $x$ satisfy $x^2 \equiv 7 \pmod{22}$, so there are no positive integers $x,y$ overall satisfy $125+x^2=22y$.

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  • $\begingroup$ Thank you for the comprehensive example. But my question now is that, are there any criteria which determines the nonexistence of solutions of given Diophantine equation? $\endgroup$
    – Jr Antalan
    May 14, 2018 at 14:52
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    $\begingroup$ Unfortunately, I did not know any other way than trial and error for these types of problems. However, you can make things easier by expressing $a$ as a multiplication of primes. You can also make $c$ smaller to make it less time-consuming to solve (like what I do above: $125=132-7=22\times 6-7$. $\endgroup$
    – user061703
    May 14, 2018 at 14:57
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Above equation shown below:

$x^2-ay+c=0$ -----------$(1)$

Equation $(1)$ has parametric solution for $(a,c)= (2,9)$

Where $(x,y)= ((2k+9), (2k^2+18k^2+45))$

For k=4, we get:

$(17) ^2-2(149)+9=0$

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